# Solving integrals using Gamma functions

• September 7th 2010, 05:45 AM
sashikanth
Solving integrals using Gamma functions
$
\int_{0}^{\infty} \cos (x^{2}) dx = \int_{0}^{\infty} \sin (x^{2}) dx = \frac{1}{2}\sqrt{\frac{\pi}{2}}
$

Can this be proved using Gamma functions without using complex numbers or De Moivre's Theorem? I'm doing problems on Gamma and Beta functions yet and I have this problem in my exercise and I haven't covered extension of integrals to complex numbers yet so I'm assuming that you can do this without using complex numbers but I fail to understand how! :(
• September 7th 2010, 11:56 PM
simplependulum
Perhaps I am driving on the detour , you will see that I am actually making use of the calculus of the parameter , together with Laplace Transform .

Let's consider $\int_0^{\infty} \sin(x^2) ~dx$ , first we make a substitution $x = \sqrt{t}$ then we have

$\frac{1}{2} ~ \int_0^{\infty} \frac{\sin(t)}{\sqrt{t}}~dt$

at this moment i remember that the Laplace Transform of $f(x) = \frac{1}{\sqrt{x}}$ is $\frac{ \sqrt{\pi} }{\sqrt{s} }$ by changing the dummy variables we have :

$\frac{1}{\sqrt{t} } = \frac{1}{\sqrt{\pi}} ~ \int_0^{\infty} e^{-ty} \frac{dy}{\sqrt{y} }$

Following it is the change of the order of integration :

$\frac{1}{2\sqrt{\pi}} ~ \int_0^{\infty} \frac{1}{\sqrt{y}} \int_0^{\infty} e^{-ty} \sin(t) ~dt~dy$

$= \frac{1}{2\sqrt{\pi}} ~ \int_0^{\infty} \frac{1}{y^2 + 1 } ~\frac{dy}{\sqrt{y}}$

Sub. $y = u^2$

$= \frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{du}{u^4 + 1 }$

Consider $I = \int_0^{\infty} \frac{u^2 - 1 }{u^4 + 1} ~du$ by changing $u = 1/t$ we have $I = - I$ so $I = 0$ . Also , the similar integral but which is not zero again :

$\int_0^{\infty} \frac{u^2 + 1 }{u^4 + 1} ~du = \int_0^{\infty} \frac{1+ 1/u^2}{ (u-1/u)^2 + 2 }~du$ , sub $u-1/u = x$

$= \int_{-\infty}^{\infty} \frac{dx}{x^2 + 2 } = \frac{\pi}{\sqrt{2}}$

By considering the difference of two integrals we have

$2 \int_0^{\infty} \frac{du}{u^4 + 1 } = \frac{\pi}{\sqrt{2}}$

or

$\int_0^{\infty} \frac{du}{u^4 + 1 } = \frac{\pi}{2\sqrt{2}}$

Finally , $\int_0^{\infty} \sin(x^2) ~dx = \frac{\pi}{2\sqrt{\pi}~\sqrt{2}} = \frac{\sqrt{\pi}}{2\sqrt{2}}$

Remarks :

If we consider the sum of these two integrals , we obtain

$\int_0^{\infty} \frac{u^2}{u^4 + 1 } ~du = \frac{\pi}{2\sqrt{2}}$ this result helps evaluate the integral $\int_0^{\infty} \cos(x^2)~dx$ which can be started in a quite similar way .
• September 8th 2010, 08:06 AM
sashikanth
Thanks! I hadn't thought of Laplace transforms at all. :)